therefore C will fall on F, or in EF or EF produced. If C falls on F, the triangles coincide, and therefore are identically equal. If C falls in EF, or EF produced, as at G, then because DG is equal to DF, therefore the angle DFG is equal to the angle DGF, therefore the angles DGE, DFE are supplementary, that is, the angles ACB, DFE are supplementary. COR. Two such triangles are identically equal Def. 30. I. 7. I. 2. Q.E.D. (1) If the two angles given equal are right angles or obtuse angles. (2) If the angles opposite to the other two equal sides are both acute, or both obtuse, or if one of them is a right angle. (3) If the side opposite the given angle in each triangle is not less than the other given side. Ex. 28. The point O is equidistant from the arms of the angle BAC, shew that OA bisects the angle BAC. EXERCISES. 29. If two quadrilaterals have three sides of the one equal respectively to three sides of the other taken in order, and have likewise the angles contained by those sides equal to one another, each to each, they are equal in all respects. 30. Two isosceles triangles are on the same base; shew that the straight line through their vertices bisects the base at right angles. 31. In the equal sides AB, AC of an isosceles triangle ABC points D and E are taken such that AD is equal to AE, if BE and CD intersect at F, shew that the triangles BFC, DFE are isosceles. *32. The three straight lines bisecting the sides of a triangle at right angles meet in a point which is equidistant from the vertices of the triangle. *33. The bisectors of the angles of a triangle meet in a point which is equidistant from the sides of the triangle. *34. The bisectors of an angle of a triangle and of the exterior angles adjacent to the other two angles meet in a point which is equidistant from the sides of the triangle. *35. The bisectors of the angles of the triangle ABC meet in O, and OF is drawn perpendicular to AB: shew that AF is equal to the difference between the semi-perimeter of the triangle and the side BC. If OD is perpendicular to BC, find similar values of BD and CD. *36. The bisector of the angle A of the triangle ABC and of the exterior angles at B and C meet in O, and OD, OE, OF are drawn perpendicular to BC, CA and AB produced when necessary: shew that AE and AF are each equal to the semi-perimeter of the triangle. Find values of BD and CD in terms of the semi-perimeter and the sides. 37. Prove that the perimeter of a triangle is greater than the sum of the straight lines drawn from the vertices to the middle points of the opposite sides. 38. If ABC is a triangle having the side AB less than the side AC, and the bisector of the angle BAC meet BC at D: shew that BD is less than CD. 39. If ABC is a triangle having the side AB less than the side AC: shew that the bisector of the angle BAC lies between AB and the straight line drawn from A to the middle point of BC. *40. Two points, A and B, lie on the same side of the straight line CD; P is a point in CD, such that AP and BP make equal angles with CD; Q is any other point in CD: shew that the sum of AP and BP is less than the sum of AQ and BQ. SECTION III. PARALLELS AND PARALLELOGRAMS. DEF. 33. Parallel straight lines are such as are in the same plane and being produced to any length both ways do not meet. DEF. 34. When a straight line intersects two other straight lines it makes with them eight angles, which have received special names in relation to the lines or to one another. 2/1 3/4 6/5 8 Thus in the figure 1, 2, 7, 8 are called exterior angles, and 3, 4, 5, 6 interior angles; again 4 and 6, 3 E and 5, are called alternate angles; lastly, I and 5, 2 and 6, 3 and 7, 4 and 8, are called corresponding angles. THEOR. 21. If a straight line intersects two other straight lines and makes the alternate angles equal, the straight lines are parallel. Let the straight line EH intersect the straight lines AB, CD at F and G so as to make the angle AFG equal to the alternate angle FGD: For if AB and CD are not parallel, they will meet if produced far enough either towards B and D or towards A and C. Suppose them to meet at a point K ; then, of the two angles AFG and FGD, the one is an exterior, I. 9. and the other an interior opposite angle, of the triangle FGK; therefore the angle AFG is not equal to the angle FGD, but by hypothesis the angle AFG is equal to the angle FGD, hence the angles AFG and FGD are both equal and unequal, which is absurd; therefore AB and CD cannot meet, that is, they are parallel. Q.E.D. Ex. 41. Shew that the contrapositive form of Theor. 21 is in cluded in Theor. 9. THEOR. 22. If a straight line intersects two other straight lines and makes either a pair of alternate angles equal, or a pair of corresponding angles equal, or a pair of interior angles on the same side supplementary; then, in each case, the two pairs of alternate angles are equal, and the four pairs of corresponding angles are equal, and the two pairs of interior angles on the same side are supplementary. Let the straight line EFGH intersect the straight lines AB, CD and make the alternate angles AFG, FGD equal: A G then shall the alternate angles BFG, FGC be equal, the cor responding angles EFB, FGD equal, and the interior angles BFG, FGD supplementary. I. 2. I. 2. Hyp. The angle BFG is the supplement of the angle AFG, the angle FGC is the supplement of the angle FGD, and the angle AFG is equal to the angle FGD, therefore the angle BFG is equal to the angle FGC. I. 1. Cor. 3. Again, the angle EFB is equal to the vertically opposite angle AFG, I. 4. |